Page:Philosophical Transactions of the Royal Society A - Volume 184.djvu/1328

These are sufficient to determine uniquely the coefficients in the matrix, and from the coefficient of $h^{9}$ we derive the differential equation to the non-singular cubic

${\frac {u_{5}^{2}\mathrm {U} _{9}-3\mathrm {U} _{7}^{2}}{u_{5}^{3}}}f+{\frac {\mathrm {V} _{8}+2u_{5}^{4}}{u_{5}}}\phi _{1}-{\frac {\mathrm {U} _{7}}{u_{5}}}\phi +3u_{5}\psi =0,$

which becomes

$u_{5}^{2}\mathrm {U} _{7}\mathrm {U} _{9}-\mathrm {V} _{8}^{2}-4\mathrm {U} _{7}^{3}+u_{5}^{4}\mathrm {V} _{8}=0,$

and is identical with that previously found in (36).

And the matrix of the non-singular cubic is

$\mathrm {U} _{7}u_{4}^{2}+{\frac {\mathrm {U} _{7}^{2}-\mathrm {V} _{8}\mathrm {L} _{6}-\mathrm {U} _{7}\mathrm {L} _{6}^{2}}{u_{5}^{2}}}u^{4}+\mathrm {V} _{8}+\mathrm {U} _{7}\mathrm {L} _{6}.$

26. The matrix of the equation to the tangents to the cubic from the temporary origin.

To find this equation, put $\pi =m\xi _{1}$, and form the discriminant of the resulting expression in $\xi$ as a binary quantic. Equating this to zero, replace $m$ by $\pi /\xi$. Since we need only the coefficient of the highest power of $\pi$, the simplest method of finding it is to put $\xi =0$ in the matrix of the cubic, and form the discriminant of the resulting expression in $\pi$.

The matrix required is

$9b_{3}^{3}-12bc_{2},$

or

$\left(\phi +\phi _{1}\mathrm {L} _{6}+\phi _{2}\mathrm {L} _{6}^{2}\right)^{2}-4f\left(\psi +\psi _{1}\mathrm {L} _{6}+psi_{2}\mathrm {L} _{6}^{2}+psi_{3}\mathrm {L} _{6}^{3}+psi_{4}\mathrm {L} _{6}^{2}\right)\quad .\quad .\quad (47),$

and the degree of this in $\mathrm {L} _{6}$ denotes the number of tangents which can be drawn.

The conditions that the cubic may be nodal or cuspidal are that this matrix, as a function of $\mathrm {L} _{6}$, may have a linear factor twice or three times repeated.

27. Case of nodal cubic.

In this case we still determine uniquely all the ratios of the coefficients, except $\psi /f$, by the condition that the coefficients of $h$ up to the seventh vanish identically, while the differential equation is found by equating the coefficient of $h^{8}$ to zero.

The further condition, to determine $\psi /f$, is (47) that the discriminant of

$\left(f\mathrm {L} _{6}^{2}+\psi \mathrm {L} _{6}-\mathrm {U} _{7}f\right)^{2}-4u_{5}^{4}f\left(f\mathrm {L} _{6}+\psi \right)$

may vanish.

Writing $k$ or $\psi /2f$, and $x$ for $\mathrm {L} _{6}+k$, this equation becomes

$x^{4}-2\left(k^{2}+\mathrm {U} _{7}\right)x^{2}-4u_{5}^{4}x+\left(k^{2}+\mathrm {U} _{7}\right)^{2}-4u_{5}^{4}k=0\quad .\quad .\quad .\quad (48).$